Problem 44 - Chemistry The Molecular Science

Question

Ten liters of \(\mathrm{F}_{2}\) gas at \(1.00 \mathrm{~atm}\) and \(100.0{ }^{\circ} \mathrm{C}\) reacts with \(99.9 \mathrm{~g} \mathrm{CaBr}_{2}\) to form \(\mathrm{CaF}_{2}\) and bromine gas. Calculate the volume of \(\mathrm{Br}_{2}\) gas formed at this temperature and pressure.

Step-by-Step Solution

Answer

Use stoichiometry based on the balanced equation and initial conditions to find the volume of \( \mathrm{Br}_{2} \) produced.

Step 1: Write the Balanced Chemical Equation

The reaction between calcium bromide and fluorine gas produces calcium fluoride and bromine gas. The balanced chemical equation is: \[ \mathrm{CaBr}_{2} + \mathrm{F}_{2} \rightarrow \mathrm{CaF}_{2} + \mathrm{Br}_{2} \]

Step 2: Calculate the Moles of Reactants

First, use the ideal gas law to find the moles of \( \mathrm{F}_{2} \): \[ PV = nRT \] Where \( P = 1.00 \mathrm{~atm} \), \( V = 10.0 \mathrm{~L} \), \( R = 0.0821 \mathrm{~L}\cdot\mathrm{atm}/(\mathrm{K} \cdot \mathrm{mol}) \), and \( T = 273.15 + 100.0 = 373.15 \mathrm{~K} \). Solving for \( n \): \[ n = \frac{PV}{RT} = \frac{1.00 \times 10.0}{0.0821 \times 373.15} \approx 0.326 \mathrm{~mol} \]Next, calculate the moles of \( \mathrm{CaBr}_{2} \):

Key Concepts

Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction where both sides of the equation have the same number of each type of atom, ensuring the law of conservation of mass is respected. In order to solve a stoichiometry problem like the one presented, it is essential to start by balancing the chemical equation. In this exercise, we are looking at a reaction between calcium bromide (\(\mathrm{CaBr}_{2}\)) and fluorine gas (\(\mathrm{F}_{2}\)) to form calcium fluoride (\(\mathrm{CaF}_{2}\)) and bromine gas (\(\mathrm{Br}_{2}\)).

The balanced equation is:
\[ \mathrm{CaBr}_{2} + \mathrm{F}_{2} \rightarrow \mathrm{CaF}_{2} + \mathrm{Br}_{2} \]
This shows that one molecule of \(\mathrm{CaBr}_{2}\) reacts with one molecule of \(\mathrm{F}_{2}\) to produce one molecule each of \(\mathrm{CaF}_{2}\) and \(\mathrm{Br}_{2}\). This equality is crucial for accurately calculating the amount of products formed in the reaction.

The balanced equation gives us a clear picture of the **molar relationships** between the reactants and products, which is the foundation for determining the volumes or masses of materials involved. Always make sure your equations are balanced before proceeding with further calculations.

Ideal Gas Law

The Ideal Gas Law is an equation of state for a hypothetical gas, and it is very useful in calculating properties of gases under various conditions. It's expressed as:

\[ PV = nRT \]
Where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant, \(0.0821 \mathrm{~L}\cdot\mathrm{atm}/(\mathrm{K} \cdot \mathrm{mol})\).
• \(T\) is the temperature in Kelvin.

In this exercise, we utilized the Ideal Gas Law to determine how many moles of fluorine gas (\(\mathrm{F}_{2}\)) were present initially. With given conditions of pressure and temperature, the Ideal Gas Law allows us to transform these measurable properties into a mole quantity.

For the reaction between \(\mathrm{CaBr}_{2}\) and \(\mathrm{F}_{2}\), knowing the moles of \(\mathrm{F}_{2}\) is essential. It helps us move forward to determine how much bromine gas \(\mathrm{Br}_{2}\) will be produced. Always remember to convert Celsius to Kelvin (\(T = 273.15 + ^{\circ}C\)) when applying this law.

The Ideal Gas Law is particularly powerful in conditions close to standard temperature and pressure, making it a go-to tool for solving gas-related stoichiometry problems.

Moles of Reactants

Understanding how many moles of each reactant you have is a vital part of solving stoichiometry problems. Moles allow chemists to relate quantities of materials in a precisely balanced chemical context.

To begin solving our exercise, we calculate the moles of \(\mathrm{F}_{2}\) using the Ideal Gas Law.
The calculation:
\[ n = \frac{PV}{RT} = \frac{1.00 \times 10.0}{0.0821 \times 373.15} \approx 0.326 \mathrm{~mol} \]
Tells us there are approximately 0.326 moles of \(\mathrm{F}_{2}\).

Next, consider the moles of \(\mathrm{CaBr}_{2}\). The molar mass of \(\mathrm{CaBr}_{2}\) is needed for this calculation. With its molecular weight, and knowing how much \(\mathrm{CaBr}_{2}\) was used, we can convert grams to moles using the equation:

• Moles = mass (g) / molar mass (g/mol)

This step ensures we know exactly how each reactant will participate in the chemical reaction.

Confirming the moles of each reactant relative to the balanced equation guides us towards understanding the stoichiometry of the reaction, thus allowing us to accurately determine the quantity of products formed.